Question: $ g(x) = \int_{-10}^{x}(10 - 3t)\,dt\,$ $ g\,'(-4)\, =$
Explanation: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = 10 - 3t$ is continuous on $[-10,-4]$. Applying the theorem We're given: $ g(x) = \int_{-10}^{x}(10 - 3t)\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =10 - 3x$ Evaluating $g'(-4)$ $ g'(-4)= 10 - 3(-4) = 22$ The answer: $g'(-4)=22$